A circuit consists of two very low-resistance rails separated by 0.300 m, with a 50.0 ohm resistor connected across them at one
1 answer:
Answer:
v = 288.67 m/s
Explanation:
Given that:
length of the separation I = 0.3
Resistance R = 50.0 ohms
Magnetic field B = 0.100 T
Power dissipated = 1.50 W
In a simple circuit, The Emf voltage of the power dissipated can be determined via the expression:
Power =
1.5 =
V = 8.660 V
From the Electromotive Force Emf ; we can calculate the speed of the bar
i.e Emf = B×v×l
8.660 = 0.1 × v × 0.3
v =
v = 288.67 m/s
You might be interested in
Answer:
54.9 m/s at 44.9 degrees
Explanation:
If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.
Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s
Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s
SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.
Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use
Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2
We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m
If we take the first derivative of the equation of position, we get the equation for speed
V(t) = Vy0 + a * t
We know that being t2 the moment the ball goes over the wall
V(t2) = -25 m/s
Y(t2) = 45.4 m
So:
45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2
-25 = Vy0 + a * t2
Then:
Vy0 = -25 - a * t2
So:
45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2
0 = -44.4 - 25 * t2 - 1/2 * a * t2^2
a = -9.81 m/s^2
0 = -44.4 - 25 * t2 + 4.9 * t2^2
Solving this quadratic equation we get:
t1 = -1.39 s
t2 = 6.5 s
Since we are looking for a positive value we disregard t1.
Now we can obtain Vy0:
Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s
Since horizontal speed is constant Vx0 = 38.9 m/s
By Pythagoras theorem we obtain the value of the initial speed:
The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90
a = atan(Vy0/Vx0) = 44.9 degrees
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>
Explanation:
The half-life of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.
In this case, we are given the half life of two elements:
beryllium-13:
beryllium-15:
As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?
We can find it out by the following expression:
Where is the amount we want to find:
Finally:
Therefore:
The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.
Answer:
0m/s²
Explanation:
Given parameters:
Initial velocity of the boat = 8m/s
Final velocity = 8m/s
Time taken = 4s
Unknown:
Acceleration of the boat = ?
Solution:
Acceleration is the rate of change of velocity with time.
It is mathematically expressed as;
A =
A is the acceleration
v is the final velocity
u is the initial velocity
t is the time taken
Insert the parameters and solve;
A = = 0m/s²